In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects ∠ABC。 If AD = 9 and DC = 7, 145–√5 is the area of a triangle.
I supposed here that [ABD] is the perimeter of ▲ ABD.
As BD is a bisector of ∠ABC ,
ABBC=ADDC=97
Let ∠B=2α
Then in isosceles △DBC
∠C=α
BC=2∗DC∗cosα=14cosα
Thus AB=18cosα
The Sum of angles in △ABC is π so
∠A=π−3α
Let's look at AC=AD+DC=16 :
AC=BCcosC+ABcosA
16=14cos2α+18cosαcos(π−3α)
[1]8=7cos2α−9cosαcos(3α)
cos(3α)=cos(α+2α)=cosαcos(2α)−sinαsin(2α)=cosα(2cos2α−1)−2cosαsin2α=cosα(4cos2α−3)
With [1]
8=cos2α(7−9(4cos2α−3))
18cos4−17cos2α+4=0
cos2α={12,49}
First root lead to α=π4 and ∠BDC=π−∠DBC−∠C=π−2α=π2 . In such case ∠A=π−∠ABD−∠ADB=π4, and △ABD is isosceles with AD=BD. As △DBC is also isosceles with BD=DC=7, AD=7≠9.
Thus first root cos2α=12 cannot be chosen and we have to stick with the second root cos2α=49. This gives cosα=23 and sinα=5√3.
The area of a triangle ABD=12h∗AD where h is the distance from B to AC.
h=BCsinC=14cosαsinα
Area of triangle ABD=145–√5
= 145–√5.
Incomplete question please read below for the proper question.
In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects ∠ABC。 If AD = 9 and DC = 7, what is the area of triangle ABD?
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